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The image of the point $(2,4)$ on the line $x+y-10=0$ is
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The correct answer is:
$(6,8)$
Let the image of point $(2,4)=(h, k)$, then $\frac{h-x_{1}}{a}=\frac{k-y_{1}}{b}=\frac{-2\left(a x_{1}+b y_{1}+c\right)}{a^{2}+b^{2}}$ $\frac{h-2}{1}=\frac{k-4}{1}=\frac{-2(2+4-10)}{1^{2}+1^{2}}$ $\frac{h-2}{1}=\frac{k-4}{1}=\frac{-2(-4)}{2}$ $\frac{h-2}{1}=\frac{k-4}{1}=4$ $h-2=4$ and $k-4=4$ $h=6$ and $k=8$
So, $(h, k)=(6,8)$
So, $(h, k)=(6,8)$
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